Trigonometry and Vectors Applied to 2D Kinematics.

Trigonometry and Vectors Applied to 2D Kinematics

There are two kinds of quantities…

Vectors are quantities that have both magnitude and direction, such as… distance, speed, mass, heat, temperature,

time…etc.

Scalars are quantities that have magnitude only, such as… displacement, velocity, acceleration, force…

etc.

Direction of Vectors Vectors are drawn as rays, or

arrows. An angle gives the direction.

Ax

Typical vector angle ranges

x

yQuadrant I0 < < 90o

Quadrant II90o < < 180o

Quadrant III180o < < 270o

Quadrant IV270o < < 360o

Direction of Vectors What angle range would this vector have? What would be the exact angle, and how

would you determine it?

Bx

Between 180o and 270o

or between-270o and -180o

Magnitude of Vectors The magnitude of a vector is the size of

whatever the vector represents. Graphically, the magnitude is represented by

the length of the vector. Symbolically, the magnitude is represented

as │A│or A

AIf vector A represents a displacement of three miles to the north…

B

Then vector B, which is twice as long, would represent a displacement of six miles to the north!

Trigonometry needed for High School Physics pythagorean theorem

sin, cos, tan (SOHCAHTOA) sin-1, cos-1, tan-1

θ

opposite

adjacent

hypotenuse

2 2 2hypotenuse adjacent opposite

A

B

RA + B = R

Graphical Addition of Vectors

R is called the resultant vector!

B

A

B

R A + B = R

The Equilibrant Vector

The vector -R is called the equilibrant. It is the inverse of the resultant. Whenever you add a vector and its inverse, you get zero.

-R

Getting components from a vector

y component

A

x

y

Ay

Ax

cos

cos

cos

x

x

adjacent

hypotenuse

A

AA A

x component

sin

sin

sin

y

y

opposite

hypotenuse

A

AA A

Getting a vector from components

angle

A

x

y

Ay

Ax

2

2 2

2 2 2

2 2

x y

x y

hypotenuse

adjacent opposite

A A A

A A A

magnitude

1

1

tan

tan

tan y

x

opposite

adjacent

opposite

adjacent

A

A

Component Addition of Vectors

1) Resolve each vector into its x- and y-components.

2) Add the x-components together to get Rx and the y-components to get Ry.

3) Calculate the magnitude of the resultant with the Pythagorean Theorem (R2 = Rx

2 + Ry2).

4) Determine the angle with the inverse tangent equation ( = tan-1 Ry/Rx.)

Vector Addition Lab

1. Attach spring scales to force board such that they all have different readings.

2. Slip graph paper between scales and board and carefully trace your set up.

3. Record readings of all three spring scales.4. Detach scales from board and remove graph paper.5. On top of your tracing, draw a force diagram by

constructing vectors proportional in length to the scale readings. Point the vectors in the direction of the forces they represent. Connect the tails of the vectors to each other in the center of the drawing.

6. On a separate sheet of graph paper, add together graphically. the three vectors

7. Then add the three vectors by component. 8. Did you get a resultant? Did you expect one? How

big is it compared to your vectors? What does the resultant represent?

Relative Motion Relative motion problems are difficult to

do unless one applies vector addition concepts.

Let’s consider a swimmer in a river. Define a vector for a swimmer’s velocity

and another vector for the velocity of the water relative to the ground. Adding those two vectors will give you the velocity of the swimmer relative to the ground.

Relative Motion

Vs

Vw

Vobs = Vs + Vw

Vw

Practice Problem: You are paddling a canoe in a river that is flowing at 4.0 mph east. You are capable of paddling at 5.0 mph.a)If you paddle east, what is your velocity relative to the shore?

b)If you paddle west, what is your velocity relative to the shore?

c)You want to paddle straight across the river, from the south to the north. At what angle to you aim your boat relative to the shore? Assume east is 0o.

Key to Solving 2-D Problems

1. Resolve all vectors into components x-component Y-component

2. Work the problem as two one-dimensional problems.

Each dimension can obey different equations of motion.

3. Re-combine the results for the two components at the end of the problem.

Sample problem: A particle passes through the origin with a speed of 6.2 m/s traveling along the y axis. If the particle accelerates in the negative x direction at 4.4 m/s2

. What are the x and y positions at 5.0 seconds?

Projectile Motion

Something is fired, thrown, shot, or hurled near the earth’s surface.

Horizontal velocity is constant and not accelerated.

Vertical velocity is accelerated. In most high school physics

classes, air resistance is ignored.

Launch Angle Launch angles can range from -90o

(throwing something straight down) to +90o (throwing something straight up) and everything in between.

90°

(up) (down)-90°

(up and out)

40° 0°

(horizontal)

Sample problem: Playing shortstop, you throw a ball horizontally to the second baseman with a speed of 22 m/s. The ball is caught by the second baseman 0.45 s later.

a) How far were you from the second baseman?

b) What is the distance of the vertical drop?

Problem: A soccer ball is kicked with a speed of 9.50 m/s at an angle of 25o above the horizontal. If the ball lands at the same level from which is was kicked, how long was it in the air?

Projectile over level ground

x

y

Range

Mathematically, the trajectory is defined by a highly symmetric parabola. The projectile spends half its time traveling up, the other half traveling down.

MaximumHeight

Acceleration of a projectile

g

g

g

g

g

x

y

Acceleration points down at 9.8 m/s2 for the entire trajectory of all projectiles.

Velocity of a projectile

vy

vx

vx

vy

vx

vy

vx

x

y

vx

vy

The velocity is tangent to path and can be resolved into components.

Position graphs for 2-D projectiles

x

y

t

y

t

x

Velocity graphs for 2-D projectiles

t

Vy

t

Vx

Acceleration graphs for 2-D projectiles

t

ay

t

ax

Derive the range equation:

R = vo2sin(2)/g.

Trigonometry and Vectors Applied to 2D Kinematics.

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